Collection of Solved Problems in Physics

A Hanging Rod

Task number: 657

• The moment of inertia

  The moment of inertia for the rotation around one of the ends of the homogenous rod of the length L, mass m is according to the table

  \[I=\frac{1}{3}mL^2. \]

• Hint

  You can use the law of conservation of mechanical energy to solve the problem. Choose two situations in which you will compare the mechanical energy of the rod. Don’t forget to choose the zero level of the potential energy.

• Hint solution

  We will compare the mechanical energy of the rod in the situation (1) in which the rod still hangs vertically down and we have just given its end speed v.

  The second situation (2) will be in the moment when the rod reaches the horizontal position and its speed equals zero.

  

  We choose the zero level of the potential energy in the middle of the rod.

  In situation (1) the rod has a zero level of potential energy. Its kinetic energy equals:

  \[E_{k1}=\frac{1}{2}I\omega^2.\]

  The total mechanical energy in situation (1) equals:

  \[E_1 = \frac{1}{2}I\omega^2.\]

  In situation (2) the rod has a zero level of kinetic energy. Its potential energy equals:

  \[E_{p2} = mg\frac{L}{2}\]

  (The whole rod is in the height L/2 above the zero level.)

  The total mechanical energy in situation (2) equals:

  \[E_2 = mg\frac{L}{2}\]

  The following holds true according to the law of the conservation of mechanical energy:

  \[E_1 = E_2\] \[\frac{1}{2}I\omega^2 = mg\frac{L}{2}\] \[\frac{1}{2}(\frac{1}{3}mL^2)\omega^2 = mg\frac{L}{2}\] \[\omega^2=\frac{3g}{L}\]

  However, we are interested in the circumferential speed, which we get from the equation v = ωr, in our case the radius of the rotation r equals length L.

  \[(\frac{v}{L})^2=\frac{3g}{L}\] \[v=\sqrt{3gL}\]

• Answer

  It is necessary to supply the end of the rod with speed \[v=\sqrt{3gL}\] .